Exercise: Find the intersections of this line with the have the same line as solution. How do you think that the equation of this plane can be specified? Exercise: What is the equation of a line through (0,0) You appear to be on a device with a "narrow" screen width (, \[a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right) + c\left( {z - {z_0}} \right) = 0\], Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities. The general equation for also satisfy the same equation. An important topic of high school algebra is "the equation of a line." In other words, if \(\vec n\) and \(\vec v\) are orthogonal then the line and the plane will be parallel. a function z of x and y. Now, if these two vectors are parallel then the line and the plane will be orthogonal. Exercise: Compare this method of finding the equation of a plane with the cross-product However, none of those equations had three variables in them and were really extensions of graphs that we could look at in two dimensions. -x - (3/2) y = -2 If c is not zero, it is often useful to think of the plane as the graph of Now, because \(\vec n\) is orthogonal to the plane, it’s also orthogonal to any vector that lies in the plane. This vector is called the normal vector. line, each of which is multiple of the other. cross product. So, the vectors aren’t parallel and so the plane and the line are not orthogonal. This is \(v = \left\langle {0, - 1,4} \right\rangle \). Also notice that we put the normal vector on the plane, but there is actually no reason to expect this to be the case. The two vectors aren’t orthogonal and so the line and plane aren’t parallel. Since both of these are in the plane any vector that is orthogonal to both of these will also be orthogonal to the plane. In general, if k is a nonzero constant, then these are equations for the Exercises: Find the equations of these lines. method. If the line is parallel to the plane then any vector parallel to the line will be orthogonal to the normal vector of the plane. We need (a) either a point on the plane and the orientation of the plane (the orientation of the plane can be specified by the orientation of the normal of the plane). root of a2 + b2 + c2. or -2 a + 5b = c. Multiply the first equation by 2 and add to eliminate a from the equation: Now, actually compute the dot product to get. with the y-axis. where at least one of the numbers a, b, c must be nonzero. for a, b, c also. This can be found expressed by determinants, or the + d(2/5)z = d, so one choice of constant gives, or another choice would be (1/5)x + (2/5)y + (2/5)z = 1. VECTOR EQUATIONS OF A PLANE. For the point F(t), we must check a[(1-t)p1+tq1] + b[(1-t)p2+tq2] This is \(\vec n = \left\langle { - 1,0,2} \right\rangle \). Thinking of a line as a geometrical object and not the graph of a function, Exercise: What is special about the equation of a plane that passes We need to find a normal vector. Recall however, that we saw how to do this in the Cross Product section. A plane is a flat, two-dimensional surface that extends infinitely far. This second form is often how we are given equations of planes. Exercise: If O is on the line, show that the equation becomes the plane that uses determinants or cross product. Sometimes it is more appropriate to utilize what is known as the vector form of the equation of plane.. Vector Form Equation of a Plane. same line, since they have the same solutions. using dot product. So, if the two vectors are parallel the line and plane will be orthogonal. and a point (h,k)? Another choice might be c = 3: x+y = 3, which So, let’s start by assuming that we know a point that is on the plane, \({P_0} = \left( {{x_0},{y_0},{z_0}} \right)\). 4b + 5b = 9b = 2c + c = 3c, so b = (1/3)c. Then substituting into the first equation for the line. This is called the scalar equation of plane. Example: P = (1, 1, 1), Q = (1, 2, 0), R = (-1, 2, 1). a line (normal form) is. Let P = (p1, p2), Q = (q1, We would like a more general equation for planes. Another useful form of the equation is to divide by |(a,b)|, Now, we know that the cross product of two vectors will be orthogonal to both of these vectors. points. Q, R all satisfy the same equation ax + by + cz = d, then all the points F(s,t) Example: For P = (1, 2), Q = (-2, 5), find the equation ax + by = c This is not as difficult a problem as it may at first appear to be. Therefore, we can use the cross product as the normal vector. For 3 points P, Q, R, the points of the plane can all be written in the parametric Often this will be written as, \[ax + by + cz = d\] where \(d = a{x_0} + b{y_0} + c{z_0}\). picture. Often this will be written as. get the equation kax + kby + kcz = kd, the plane of solutions is the same. Given the coordinates of P, Q, R, there is a formula for the coefficients of of line PQ. 4x + 6y = 8 A plane in 3-space has the equation . axes? This section is solely concerned with planes embedded in three dimensions: specifically, in R . Line through (3, 4) and (-6, -8). Exercise: Where does the plane ax + by + cz = d intersect the coordinate If the coordinates of P and Q are known, then the coefficients a, b, c of an 1, -1), and (1, -1, -1)? A normal vector is. ax + by = 0, or y = mx. Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane. Exercise. Finally, since we are going to be working with vectors initially we’ll let \(\overrightarrow {{r_0}} \) and \(\vec r\) be the position vectors for P0
What is equation of the plane through the points I, J, K? Given points P, Q, R in space, find the equation of the plane through the 3 Notice as well that there are many possible vectors to use here, we just chose two of the possibilities. be converted to slope-intercept form by solving for y: except for the special case b = 0, when the line is parallel to the y-axis. This choice will of a plane (when at least one of a, b, c is not zero. Line through (3, 4) and (1, -2). q2). A plane is the two-dimensional analog of a point (zero dimensions), a line (one dimension), and three-dimensional space. Both, Vector and Cartesian equations of a plane in normal form are covered and explained in simple terms for your understanding. be explained in the Normal Vector section. Then since the points are on the line, we know that both. If both P and Q satisfy the same Notice that we added in the vector \(\vec r - \overrightarrow {{r_0}} \) which will lie completely in the plane. This means an equation in x and y whose solution set is a line in the (x,y) This computation will not be done here, since it can be done much more simply Now, assume that\(P = \left( {x,y,z} \right)\) is any point in the plane. We put it here to illustrate the point. plane. Line through (3, 4) and (3, 7). This in effect uses x as a parameter and writes y as a function of x: y = f(x) We can factor out c (or set c equation holds. These two vectors will lie completely in the plane since we formed them from points that were in the plane. = 1 for the same result) and get (1/3)x + (1/3)y =1 as one choice of (b) or a point on the plane and two vectors coplanar with the plane. (1/2)x + (3/4)y = 1. Equation of a Plane. through 0. Start with the first form of the vector equation and write down a vector for the difference. The computations are the same, but one We can pick off a vector that is normal to the plane. Learn to derive the equation of a plane in normal form through this lesson. This can easily Now, let’s check to see if the plane and line are parallel. As for the line, if the equation is multiplied by any nonzero constant k to get the equation kax + kby + kcz = kd, the plane of solutions is the same. + tQ, for any choice of t. Here is this computation. equation ax+by = c, then a computation shows that this is also true for (1-t)P As for the line, if the equation is multiplied by any nonzero constant k to Adding the equations gives 5b = 2d, or b = (2/5)d, then solving for c = b = We can form the following two vectors from the given points. (1/c). If the coefficients on the normal form are multiplied by a nonzero constant, in the Normal Vector section. or a + 2b = c. 2x + 3 y = 4 This is called the vector equation of the plane. a + 2b + 0c = d We can also get a vector that is parallel to the line. equation, a = c - 2b = c - (2/3)c = (1/3)c. This gives the equation [(1/3)c]x + [(1/3)c}y = c. Why is the c not Compare this explicit computation with the computation given If \(\vec n\) and \(\vec v\) are parallel, then \(\vec v\) is orthogonal to the plane, but \(\vec v\) is also parallel to the line. In particular it’s orthogonal to \(\vec r - \overrightarrow {{r_0}} \). term = 1. This is called the scalar equation of plane. to the third, we eliminate a to get. has cleared the denominators. of an equation ax + by + cz = d, where P, Q and R satisfy the equations, thus: a + b + c = d In the first section of this chapter we saw a couple of equations of planes. = mx+b. shows more detail and one hides the coordinates and shows a more conceptual Another useful form of the equation is to divide by |(a,b,c)|, the square form F(s,t) = (1 - s - t)P + sQ + tR, where s and t range over all real numbers. The equation can be rearranged like this: Another useful choice, when d is not zero, is to divide by d so that the constant + t(aq1 + bq2), and this equals (1-t)c + tc = c. So the It is completely possible that the normal vector does not touch the plane in any way. Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane. Note the special cases. If you think about it this makes some sense. for the plane that uses dot product. coordinate axes. -a + 2b + c = d, Subtracting the first equation from the second and then adding the first equation Given two points P and Q, the points of line PQ can be written as F(t) = (1-t)P the square root of a2 + b2. Since P is on the line, its coordinates satisfy the equation: a1 + b2 = c, In order to write down the equation of plane we need a point (we’ve got three so we’re cool there) and a normal vector. = c. But the left side can be rearranged as (1-t)(ap1 + bp2) the set of solutions is exactly the same, so, for example, all these equations So, the line and the plane are neither orthogonal nor parallel. 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